CS290A, Spring 2005: & Quantum Computation

Wim van Dam Engineering 1, Room 5109 vandam@cs

http://www.cs.ucsb.edu/~vandam/teaching/CS290/ Administrative

• The Final Examination will be: Monday June 6, 12:00–15:00, PHELPS 1401

• New Exercises are posted Try to answer Question 2 before Thursday. Last Week / This Week

• Last week we looked at quantum money and , which uses the states 0,1,+,–.

• This week we will extend this idea to describe “quantum fingerprinting”.

• Also this week: superdense quantum coding and of quantum states. Fingerprinting

Assume two parties A and B that each have data in the form of a (long) string x and y ∈{0,1} N.

A and B want to check if they have the same data, without revealing a priori to the other their strings. They do this by sending (publicly) information about their strings (x and y) to a trusted third party C, who decides.

Sending the whole strings is not allowed because the strings are too long / risk of eavesdropping.

A and B want to have a way of fingerprinting their strings. Quantum Fingerprinting

“Quantum Fingerprinting” refers to a way of mapping the ∈ N strings x,y {0,1} to quantum states | φx〉 and | φy〉, that live in a ‘much smaller than 2 N’-dimensional Hilbert space, such that from ψ and φ we can tell decide whether x=y.

∈ N The set { |φx〉 : x {0,1} } cannot be mutually orthogonal. Instead we will have to work with near orthogonal states.

∈ N Central Idea: Encode x {0,1} into m qubit state |φx〉 (with m much smaller than N). 2 Do this in a way such that | 〈φx|φy〉| ≤ ½ if x≠y. Third party decides if |φx〉 = |φy〉 or not. Simple Example

Let x and y be from a set of 6 possibilities.

Let | φx〉 and |φy〉 be from the set of 6 states {|0〉, |1〉, (|0〉+|1〉)/ √2, (|0〉–|1〉)/ √2, (|0〉+i|1〉)/ √2, (|0〉–i|1〉)/ √2}

2 For all qubit states with x≠y we have |〈φx|φy〉| ≤ ½.

The third party receives two unknown states |φx〉 and |φy〉 that are either the same or very different.

How to distinguish between these two possibilities?

A Equality Tester for unknown states can be implemented with a Controlled Swap Test… Controlled Swap Test

• Given two unknown quantum states | ψ〉 and | φ〉, are they the same or not? • You can test this using the Controlled Swap Test” C-SWAP:|0,x,y 〉 # |0,x,y 〉 |0 |0 ? C-SWAP:|1,x,y 〉 # |1,y,x 〉 〉 H H 〉 in the circuit……………. | S φ〉 W | A ψ〉 P

• Observing a “0” indicates that |ψ〉 and | φ〉 are close to each other, observing a “1” that they are far apart.

• What are the exact probabilities? Probabilities of C-SWAP

• The evolution of the system is |0 〉 H H |0 〉? ,0 , a 1 ( ,0 , + ,1 , ) | S φ ψ 2 φ ψ φ ψ φ〉 W A 1 | 〉 a ( ,0 , + ,1 , ) ψ P 2 φ ψ ψ φ a 1 + + − 2 ( ,0 φ,ψ ,1φ,ψ ,0 ψ,φ ,1ψ,φ ) = ⊗ 1 + + ⊗ 1 − 0 2 ( φ,ψ ψ,φ ) 1 2 ( φ,ψ ψ,φ ) The probability of observing a “0” is therefore = 1 + + Prob("0" ) 4 ( φ,ψ ψ,φ )( φ,ψ ψ,φ ) = 1 + + 4 2( ψ,φ φ,ψ φ,ψ ψ,φ ) 2 = 1 + 1 2 2 ψ φ Equality Testing

• The previous calculations |0 〉 H H |0 〉? show that if | , |2 1, 〈φ ψ〉 ≈ |φ〉 S then the probability of W |ψ〉 A observing a “0” is ≈ 1. P • If | 〈φ,ψ〉|2 ≈ 0, then the probability of “0” is ≈ ½.

• By repeating the experiment a number of times (using fresh copies of | ψ〉 and | φ〉), we can –with near certainty– distinguish between the cases |ψ〉 = | φ〉 and |〈φ,ψ〉|2 ≤ ½. More on Q-Fingerprinting

Using methods from error correction it is possible to encode N bits of information into M ≈ c log N qubits 2 such that |〈φx|φy〉| ≤ ½ if x≠y.

Even if we have to send many copies of the the fingerprint, it will still be more efficient than sending the N classical bits of the original strings x and y.

Added advantage: because we send such a highly compressed quantum state, it is impossible to infer the string x from the fingerprint |φx〉. Communication & Entanglement

• Holevo’s bound tells us that we can encode only one bit of information into one qubit. This bound assumes that the qubit is unentangled.

• Things change if we allow the communication of qubits that are entangled with qubits of the receiver.

• What happens if A and B share a prior entangled

pair of qubits |EPR 〉 = (|0 A,0 B〉+|1 A,1 B〉)/ √2 before Alice is to send (quantum) information to Bob? Superdense Quantum Coding

• Let A and B share an EPR pair of qubits. Alice wants to transmit classical information to Bob.

• Using , A can send two bits of information to B, using only one qubit.

• Approach: Depending on A’s input {1,2,3,4} she applies to her side of the EPR pair, one of 4 transformations { I, X, Y, Z}, and sends her EPR qubit to Bob. • From the changed EPR pair, Bob decodes which transformation A has applied. This can be done reliably and will transfer 2 bits of information from A to B. How to Do It?

• Look back at Question 1, Exercises 2 (Week 3)

)I( a 1 ( 00 + 11 ) • The four transformations 2 give us four entangled (X) a 1 ( 10 + 01 ) states that are mutually 2 orthogonal… (Y) a 1 10i( − i 01 ) 2 )Z( a 1 ( 00 − 11 ) 2 • With a CNOT and a Hadamard, Bob can map those states to the outputs |00 〉, |01 〉, –i|11 〉 |and |10 〉. More on Superdense Coding

• One can prove that the 2 bits / 1 qubit ratio is optimal: it is not possible to send (say) 3 bits of information using 1 qubit and lots of entanglement.

• Superdense coding is not possible classically.

• Experimental implementation of superdense coding: [Zeilinger et al.,1996, Innsbruck, Austria]

• Can we do the inverse: Send quantum information using classical communication between A and B? What is Teleportation?

Alice unknown qubit Bob q = α 0 + β 1

Alice wants to send her unknown quantum information to Bob.

A and B do not have a : se n: U utio ed only classical communication is allowed. ol ngl S nta n e ir a pa Alice cannot tell Bob what the values α,β are, EPR nor can she measure |q 〉 to see what they are. Towards Teleportation

The first two qubits are on Alice’s side The 3 rd one is Bob’s

After H/CNOT, they share an EPR pair.

Alice performs a CNOT and a Hadamard to her 2 qubits and measures them in the 0/1 basis.

What can she expect to observe? Output?

q ⊗ EPR = ( 0 + 1 ) ⊗ 1 ( 00 + 11 ) α β 2 = 1 ( 00,0 + 11,0 + 00,1 + 11,1 ) 2 α α β β a 1 ( 00,0 + 11,0 + 10,1 + 01,1 ) 2 α α β β a 1 + + + + 2 (α 00,0 α 00,1 α 11,0 α 11,1 β 10,0 − β 10,1 + β 01,0 − β 01,1 ) Regardless of = 1 ⊗ + + 2 00 (α 0 β 1 ) the values α,β 1 01 ⊗ ( 1 + 0 ) + the probability 2 α β of measuring 1 ⊗ − + 2 10 (α 0 β 1 ) {00,01,10,11} 1 ⊗ − is one-quarter. 2 11 (α 1 β 0 ) Effect of A’s Measurement

= 1 00 ⊗ ( 0 + 1 ) + 2 α β Depending on the two 1 ⊗ + + leftmost qubits (A’s side), 2 01 (α 1 β 0 ) the third qubit (B’s side) 1 ⊗ − + 2 10 (α 0 β 1 ) is 1-out of-4 permutations 1 ⊗ − of the original qubit |q . 2 11 (α 1 β 0 ) 〉

When A measures the Outcome “00”: → α|0 〉+β|1 〉 two bits ∈{00,01,10,11} Outcome “01”: → α|1 〉+β|0 〉 the qubit of B collapses Outcome “10”: → α|0 〉–β|1 〉 to 1 of the 4 versions of q. Outcome “11”: → α|1 〉–β|0 〉

When A tells B which one of the 4 outcomes she has observed, Bob knows what to do to correct his qubit to the original |q 〉. Bob’s Correction

“00”: → α|0 〉+β|1 〉 When A tells B which one of the “01”: → α|1 〉+β|0 〉 4 outcomes she has observed, “10”: → α|0 〉–β|1 〉 Bob knows what to do to correct “11”: → α|1 〉–β|0 〉 his qubit to the original |q 〉…

If first bit is a “1”, Bob applies a Z gate |b 〉 # (–1) b|b 〉. If 2nd bit is a “1”, Bob applies a NOT |b 〉 # |b ⊕1〉

The outcome will always be α|0 〉+β|1 〉 = |q 〉

Bob has (re)created the unknown qubit q that was on Alice’s side. During the process Alice has ‘lost’ her copy of the qubit (otherwise we would have copied q). Teleportation Alice Shared EPR pair Bob unknown qubit |q 〉 = α|0 〉+β|1 〉.

G Alice applies a CNOT and a Hadamard to Bob receives the two q and her EPR half. bits and acts on his side of the EPR, in She measures the bits in a way that recreates the 0/1 basis and sends the unknown qubit the information to Bob. |q 〉 = α|0 〉+β|1 〉. What Just Happened?

• Teleportation requires one EPR pair and two classical bits to transfer one qubit from A to B. • Superdense coding requires one EPR pair and one quantum bit to transfer two classical bits of information.

• The qubit did not get copied: Alice’s measurement destroyed it on her side.

• The outcome of Alice’s measurement is completely random and independent of the α,β values of |q 〉. A and B learn nothing about the unknown qubit. Alice’s Measurement

• Initially, Bob’s part of the EPR pair has nothing to do with the qubit q on Alice’s side. • After she has performed her measurement, this appears to have changed: Now Bob’s EPR-half is (almost) identical to q (except for some ‘corrections’). • Does Alice’s measurement instantaneously change B’s qubit in a way that can be used for communication?

• Answer: No (of course). Until Bob has received the two bits of information from Alice, his qubit remains as random as it was before the measurement. So What does Happen?

• What exactly happens when one half of an EPR pair gets measured is a deep question in physics.

• Example, take |EPR 〉=(|00 〉+|11 〉)/ √2 between A and B. • If B measures his part, he will see 0/1 with 50%/50%. We say that this is caused by of the unavoidable randomness of quantum physics. • But if A has measured her qubit beforehand, then she knows with 100% that Bob will observe the same value.

• When is the outcome of a measurement determined? Classical Correlations

• Compare a classically correlated state: Two distributed bits with are promised to be equal, but are otherwise random (“00” or “11”)

• Again, Bob does not know what value he will see. But when Alice knows her value, she can predict Bob’s value with 100% accuracy. • In this situation we would say that the randomness on Bob’s side is due to his ignorance : The outcome is predetermined, but he is just unaware of it.

• In quantum mechanics, on the other hand, the outcome does not seem to be predetermined… Complete Description?

• The question is: When we say that two qubits are in the state |EPR 〉=(|00 〉+|11 〉)/ √2, is that all there is to know? • From it, we cannot predict what the outcomes of our measurements will be, so some information seems to be missing in our description (ignorance).

• Are there “hidden variables” that we could include in our description of the quantum state that would predict the outcomes of (future) measurements?

• Answer: This is not the case. Hidden Variables

In a hidden variable theory, the particles A and B have determined beforehand what the outcomes will be when (later on) they are measured.

Several kinds of measurements are possible so each particle would have a list of answers for the various kinds of measurements: imic not m ts can tum Measurement : Outcome : uch lis quan S tics of in statis e see “0”? “Yes” the that w hanics “+”? “No” mec ry. borato the la Three Party Entanglement

Greenberger, Horne and Zeilinger described a three qubit state that is entangled over three parties A,B,C: |GHZ 〉=(|000 〉+|111 〉)/ √2

Party A will change the phase of her qubit according to α |0 〉#|0 〉 and |1 〉#ei |1 〉. Same for B,C with angles β,γ. After that they perform a Hadamard gate and measure their qubits in the standard 0/1 basis… Midterm Flashback

• What happens was the last question on the Midterm.

• If the sum α+β+γ = 0 mod 2 π then the parity of the outcome bits will be even. If α+β+γ = π mod 2 π, then the parity of the three bits will be odd.

• You can use this to implement a distributed even/odd deciding algorithm that minimizes the communication between the three parties A,B and C. • It is impossible to implement this algorithm using a “hidden variables technique”…